设z=yf(x2-y2),其中f(u)可微,则=____________.

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问题 设z=yf(x2-y2),其中f(u)可微,则=____________.

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答案[*]

解析 =yf′(x2一y2).2x=2xyf′(x2一y2),
=yf′(x2一y2)(一2y)+f(x2一y2)=-2y2f′(x2一y2)+f(x2一y2),
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